How can I calculate $\lim_{(\alpha,\beta)\to(0,0)}\frac{f(\alpha,\beta)}{g(\alpha,\beta)}$?

Question

What's the value of $$ \lim_{(\alpha,\beta)\to(0,0)}\frac{[\lambda_i\alpha+\mu_i\beta+O(r^2)] [\lambda_j\alpha+\mu_j\beta+O(r^2)][-\lambda\alpha^2-\mu\beta^2+O(r^3)]} {\left(\sum_{i=1}^3\left(\lambda_i\alpha+\mu_i\beta+O(r^2)\right)^2\right)^{3/2}},\quad i,j = 1,2,3\tag{*} $$ where $\lambda_i,\mu_i,\lambda,\mu$ are constants, $r = \sqrt{\alpha^2+\beta^2}$?

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I guess this limit may not exist, OR if it does exist, it should be $0$. For investigating the possibility, I tried some simpler case. For instance, let $\lambda_i=\mu_i=-\lambda=-\mu=1$ for $i=1,2,3$ and $O(r^3)=O(r^2)=0$. Then the limit becomes $$ \lim_{(\alpha,\beta)\to(0,0)}\frac{(\alpha+\beta)^2(\alpha^2+\beta^2)}{3^{3/2}(\alpha+\beta)^3}. $$

How should I deal with the term such as $$ \lim_{(\alpha,\beta)\to(0,0)}\frac{\alpha^4}{(\alpha+\beta)^3} ? $$ In general what's the value of (*)?

For the polynomial with only one variable, we have $$ \lim_{x\to 0}\frac{f(x)}{g(x)}=0 $$ when the order of $f$ is higher than that of $g$.

Do we have the same result for the multivariate polynomials which appears in the title?

Answer 1

In light of the answer and comments to this question, I think the polar coordinates can help. Let $$\alpha=r\cos\theta,\quad\beta=r\sin\theta$$(*) can finally be reduced to $$ \lim_{r\to 0}\frac{f(r)}{g(r)}Q(\sin\theta,\cos\theta,r) $$ where $Q$ is bounded, and $f,g$ are polynomials with respect to $r$.